Answer: 11. Solution. Let in triangle $ABC$ the angle $\alpha$ be equal to the difference of angles $\beta-\gamma$. Since $\alpha+\beta+\gamma=180^{\circ}$, then $\beta-\gamma+\beta+\gamma=180^{\circ}$ and therefore $\beta=90^{\circ}$. If $\beta=90^{\circ}$ is twice one of the angles $\alpha$ or $\gamma$, then the triangle will be isosceles, which contradicts the condition. Let for definiteness $\alpha>\gamma$. Then $\beta=90^{\circ}, \alpha=60^{\circ}, \gamma=30^{\circ}$.

By the inscribed angle theorem:

$\angle MLC = \angle MBC = 45^{\circ}, \angle CLO = \angle CAO = 30^{\circ}$. Therefore, $\angle MLO = 75^{\circ}$. Similarly,

$\angle LOM = \angle LOA + \angle AOM = \angle LCA + \angle ABM = 15^{\circ} + 45^{\circ} = 60^{\circ}$,

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$\angle OML = \angle OMB + \angle BML = \angle OAB + \angle BCL = 30^{\circ} + 15^{\circ} = 45^{\circ}$.

Thus, the angles of triangle $LOM$ are $\alpha_{1}=75^{\circ}, \beta_{1}=60^{\circ}, \gamma_{1}=45^{\circ}$.

Using the sine rule, we get the formula for the area of a triangle inscribed in a circle of radius $R: S=\frac{1}{2} a b \sin \gamma=\frac{1}{2} \cdot 2 R \sin \alpha \cdot 2 R \sin \beta \cdot \sin \gamma=2 R^{2} \sin \alpha \sin \beta \sin \gamma$.

Therefore, $\frac{S_{LOM}}{S_{ABC}}=\frac{2 R \cdot \sin \alpha_{1} \sin \beta_{1} \sin \gamma_{1}}{2 R \cdot \sin \alpha \sin \beta \sin \gamma}=\frac{\sin 75^{\circ} \sin 60^{\circ} \sin 45^{\circ}}{\sin 90^{\circ} \sin 60^{\circ} \sin 30^{\circ}}=\sqrt{2} \cdot \sin 75^{\circ}$. Since $\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)=\frac{\sqrt{6}+\sqrt{2}}{4}$, we get: $\frac{S_{LOM}}{S_{ABC}}=\frac{\sqrt{3}+1}{2}$.

Therefore, $S_{LOM}=4 \sqrt{3}+4 \approx 11$.